course phy 121 10/27 9 pm Class 091021short-answer questions
Class 091026
short-answer questions
`q001. ** Give your answers to the following one-step questions:
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The KE of a 10 kg object moving at 5 m/s is 1/2 m v^2 = 1/2 *
10 kg * (5 m/s)^2 = 5 kg * 25 m^2 / s^2 = 125 kg m^2/s^2 = 125 J.`dPE of a system is equal and opposite to work done by a
conservative force acting on the system. The conservative force here is gravity, which exerts a downward force of 10 kg * 9.8 m/s^2 = 98 N as the object moves 3 meters in the upward direction. Thus gravity does negative work:`dW_grav_ON = -98 N * 3 m = -294 J
and the PE change is equal and opposite
`dPE = -`dW_grav_ON = - (-294 J) = 294 J.
The force increases linearly, so the average force is the
average of the initial and final forces, in this case (4 N + 16 N) / 2 = 10 N. The displacement is 15 cm opposite the direction of the force (the object moves one way, the conservative acting ON the object force pulls back the other way), so conservative force does work -10 N * .15 m = -1.5 N * m = -1.5 J.If `dKE = +8 J and `dPE = -12 J, then
dW_NC_ON = `dKE+ `dPE = +8 J + (-12 J) = -4 J.
`dW_NC_ON = `dKE + `dPE, so `dPE = dW_NC_ON - `dKE = 20 J - 15
J = 5 J.&&&&
`q002. ** A slingshot shoots a pebble in the upward direction. While the slingshot is in contact with the pebble:
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The elastic force exerted by the slingshot as it 'snaps back'
toward equilibrium is in the upward direction, as is the displacement of the pebble.The elastic force therefore does positive work on the pebble.
It follows that the change in elastic PE of the system is
negative, so the elastic PE decreases.The gravitational force acts downward while the pebble rises.
The gravitational force therefore does negative work on the
pebble.Its gravitational PE therefore decreases.
The slingshot manages to 'shoot' the pebble upwards,
increasing its KE. Whether we assume that `dW_NC_ON is zero or negative, we conclude that `dPE = `dW_NC_ON - `dKE must be negative.&&&&
`q003. ** For the ball rolling in two directions along the 'constant-velocity' incline (i.e., an incline on which the velocity of the ball as it rolls down the incline remains unchanged), as you observed it in class:
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Suppose that the ball traveled 40 cm 'up' the incline, requiring 8.5 half-swings of a 16-cm pendulum to come to rest (traveling 'down' the incline, presumably, it has zero acceleration and if starts does not come to rest). Then the period of the pendulum is .2 sqrt(16) s = .8 s, and 8.5 half-swings require 6.8 sec. The average velocity on the incline is
v_Ave = `ds / `dt = 40 cm / (6.8 s) = 6 cm/s, approx..
Since final velocity is zero, assuming acceleration to be
constant we see that the initial velocity must be double the average velocity, sov0 = 2 * 6 cm/s = 12 cm/s.
The change in velocity is therefore `dv = -12 cm/s, and
the acceleration isa = `dv / `dt = 12 cm/s / (6.8 s) = 1.7 cm/s^2, approx..
The difference between the 0 acceleration 'down' the ramp
and the 1.7 cm/s^2 acceleration 'up' the ramp is due to friction. The ramp has sufficient slope that the 'parallel component' of the gravitational force exactly counters the frictional force. When the ball travels 'up' the ramp it encounters the 'parallel component' of the gravitational force, which now resists its motion, as well as the frictional force. These forces are equal, so each alone would produce an acceleration of magnitude 1/2 * 1.7 cm/s^2 = .85 cm/s^2.If the ball was rolled along a perfectly level ramp, the
'parallel component' of its acceleration would be zero, and the net force would be just the frictional force. Its acceleration would therefore be .85 cm/s^2, in the direction opposite its motion. Rolled in opposite directions, then, its accelerations might therefore be -.85 cm/s^2 and +.85 cm/s^2. The difference in the accelerations would still be 1.7 cm/s^2, as was the case for the constant-velocity ramp.&&&&
`q004. The rolling friction between a steel ball and a steel incline exerts a force opposed to the direction of motion. If this force is equal to .3% of the weight of the rolling ball, then
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The weight of the ball is the force exerted on it by gravity, which is the product of its mass and the acceleration of gravity. The magnitude of the weight is therefore
weight = m g = .060 kg * 9.8 m/s^2 = .6 N, approx.
The frictional force is .3% of this, or
f_frict = .003 * .6 N = .0018 N.
On a perfectly level ramp, the net force will be the frictional force, so the acceleration of the ball will be
a = F_net / m = .0018 N / (.060 kg) = .03 m/s^2.
On a 1% incline, the gravitational force will make an angle of 271 deg, provided the positive x axis is oriented down the ramp. The x component of the gravitational force (which is parallel to the incline, hence the name 'parallel component') is therefore
x comp of wt = wt * cos(271 deg) = .6 N * .017 = .011 N, approx.
and the net force on the ball is
F_net = wt_x + f_frict = .011 N + (-.0018 N) = .009 N, where the positive direction is taken to be down the incline (the ball accelerates in the positive direction, friction being opposite the motion down the ramp is negative)
Its acceleration will therefore be
a = F_net / m = .009 N / (.060 kg) = .15 m/s^2, down the incline.
Moving up the incline the parallel component of the gravitational force and the frictional force are both down the incline, so keeping the earlier choice of positive direction the net force is
F_net = .011 N + .0018 N = .013 N, approx.
and the acceleration is
a = F_net / m = .013 N / (.060 kg) = ,22 m/s^2, approx..
So a ball travcling down the ramp (in the direction we have chosen as positive)( experiences net force .009 N down the ramp, and accelerates at .15 m/s^2 down the incline, which speeds it up. A ball traveling up the ramp (now in our chosen negative direction) experiences a net force of .013 N and an acceleration of .22 m/s^2, which slows it down.
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`q005. The ball in the preceding rolls 50 cm down the 1 degree ramp, subject to only gravity, normal force and rolling friction. For this roll:
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Rolling 50 cm down the ramp with the 'parallel' gravitational force wt_x = .011 N, the gravitational force does work
`dW_grav_ON = wt_x * ds_x = ,011 N * .050 m = .00055 J
on the ball, resulting in
`dPE = - `dW_grav_ON = -.00055 J.
The nonconservative force is the frictional force, which is .0018 N in the direction opposite motion, so it does negative work:
`dW_NC_ON = -.0018 N * .050 m = -.00009 J.
Its KE change is therefore
`dKE = `dW_NC_ON - `dPE = -.00009 J - (-.00055 J) = .00046 J
If the ball starts from rest, then its initial KE is 0 so its final KE is
KE_f = KE_0 + `dKE = 0 + .00046 J = .00046 J.
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`q006. The acceleration of a 60-gram steel ball on a certain ramp with a small incline is observed to be 2.5 cm/s^2 when the ball moves in one direction, and -1.5 cm/s^2 when the ball moves in the opposite direction.
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Moving in the positive direction the acceleration is 2.5 cm/s^2, so
F_net = .060 kg * .025 m/s^2 = .0015 N, approx.
Moving in the negative direction the acceleration is -1.5 cm/s^2, so
F_net = .060 kg * (-.015 m/s^2) = -.0009 N, approx..
The net force differs because friction acts in opposite directions when the ball moves in opposite directions. All other forces are the same. So the difference in the two forces is double the force of friction.
The difference between the two forces is .0015 N - (.0009 N) = .0024 N, so the frictional force is half this:
f_frict = difference between forces / 2 = .0024 N / 2 = .0012 N.
The average of the two forces is (.0015 N + (-.0009 N) / 2 = .0003 N.
This is the 'parallel component' of the weight, i.e.
wt_x = .0003 N.
The weight of the ball is .06 N.
You can use the Pythagorean Theorem to find the y component of the weight; since the x component is much less than the weight, the y component will differ very slightly from the .06 N weight; the y component of the weight will of course be negative.
So the weight vector has x component .0003 N and y component .06 N. Its angle with the positive x axis is therefore
angle = arcTan ( .06 N / .0003 N) = arcTan(200) = -89.7 degrees, approx..
Thus the weight vector lies .3 degrees from the negative y axis, so the incline lies .3 degrees above horizontal.
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`q007. ** My mass is about 75 kg. In about 40 minutes, my altitude and velocity change as follows:
This scenario naturally breaks into five intervals, one occurring on the first runway, another between the end of that runway and cruising altitude, a third between cruising altitude and the beginning of descent, the fourth between the beginning of the descent and first contact with the second runway, and the last on the second runway.
Give quantitative answers to each of the following:
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During the first phase, altitude does not change so gravitational PE does not change; i.e., `dPE = 0. KE changes from 0 to 1/2 * 75 kg * (50 m/s)^2 = 9400 J, approx. The nonconservative forces acting are air resistance, which acts in the direction opposite motion and therefore does negative work, rolling friction, which similarly does negative work, and the force exerted on the airplane by the air expelled from the jet engines. The latter is equal and opposite to the force exerted by the engines on the air. The air is expelled toward the back of the aircraft, which requires a backward force. The force of the expelled air on the engines is therefore forward, in the direction of motion, and does positive work on the airplane.
During this phase `dW_NC_ON = `dKE + `dPE = 9400 J + 0 J = 9400 J. This is the sum of the positive work done in reaction to the force exerted by the engines, and the negative work done by air resistance. In other words, the engines have to supply the 9400 J to accelerate my mass, in addition to my share of the air resistance.
During the second phase, KE changes from about 9400 J to 1/2 * 75 kg * (200 m/s)^2 = 1 500 000 J and gravitational PE changes by 75 kg * 9.8 m/s^2 * 10 000 m = 7 500 000 J, approx.. The nonconservative forces are air resistance and the force exerted by the expelled air on the engines.
During this phase `dW_NC_ON = `dKE + `dPE = 1 500 000 J + 7 500 000 J = 9 000 000 J. This energy, plus the energy required to overcome air resistance, are supplied by the fuel that runs the engines.
During the third phase neither the speed of the airplane nor its altitude change, so neither KE nor PE change.
During this phase `dW_NC_ON = `dPE + `dKE = 0. The negative work done by the air resistance and the positive work done by the air expelled by the engines are equal and opposite.
During the fourth phase the PE changes by -7 500 000 J, approx., and KE decreases to 1/2 * 75 kg * (60 m/s)^2 = 13 500 J.
During this phase `dW_NC_ON = `dPE + `dKE = -7 500 000 J - 1 500 000 J = -9 000 000 J. This is the sum of the negative work done by air resistance and the positive work done by the engines. To get to the runway with a survivable speed, the airplane must position itself in such a way as to intercept a lot of air. Remember, that -9 000 000 J is just for my mass.
Between contacting the runway and coming to rest, PE does not change and KE decreases from 13 500 J to 0.
Thus `dW_NC_ON = `dPE + `dKE = 0 - 13 500 J = -13 500 J. Air resistance won't accomplish this; the airplane relies on brakes, friction and reversing its engines to come to rest.
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`q008. A certain Atwood machine has 50 paperclips on each side. The paperclips have reasonably uniform mass. A positive direction has been declared, and the following observations are made.
If one of the clips is transferred to the positive side the system remains at rest.
If another clip is transferred to the positive side the system still remains at rest.
If a third clip is transferred to the positive side, the system accelerates at 15 cm/s^2.
After restoring the system to its original state, the system remains at rest after a transfer of 2 clips to the negative side.
A third clip transferred to the negative side results in an acceleration of - 10 cm/s^2.
First answer the following:
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The system has mass 100 * m_clip, where m_clip is the mass of a single clip.
With 50 clips on each side, the positive and negative forces are equal, and the net force is zero. The system does not accelerate.
With 51 clips on one side and 49 clips on the other, the downward forces on the two masses are 51 * m_clip * g and 49 * m_clip * g. One force pulls the system in the positive direction and the other in the negative direction, so the net force has magnitude
| F_net | = 51 m_clip * g - 49 * m_clip * g = 2 m_clip * g.
That is, the net force is equal to the force exerted by gravity on 2 clips.
The magnitude of the acceleration is
| a | = | F_net | / m_system = 2 m_clip * g / (100 m_clip * g) = .02 * g.
That is, the force exerted by gravity on 2 clips will accelerate a system with a mass equal to that of 100 clips at 2/100 the acceleration of gravity.
.02 * g = .196 m/s^2 (easily calculated from g = 9.8 m/s^2).
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Now, in terms of the stated data:
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`q009. ** A 40 kg block rests on an incline which makes an angle of 37 degrees with the horizontal. The coefficient of friction between the block and the incline is mu = .5. Sketch a free-body diagram for this block, sketch x and y axes with the x axis parallel to the ramp, and estimate the x and y components of the weight of the block. Then using the sine and cosine functions, calculate these components.
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Your sketch should depict the x-y axes rotated 37 degrees, either clockwise or counterclockwise, from the 'standard' vertical-horizontal orientation. The vertical weight vector will therefore lie 37 degrees from the negative x axis.
Assuming counterclockwise rotation, with the ramp inclined up and to the right, the weight vector will lie in the third quadrant, 37 degrees from the negative y axis, at angle 270 deg - 37 deg = 233 deg counterclockwise from the positive x axis. (If your ramp is inclined down and to the right the rotation of the axes will be clockwise and the angle will be 270 deg + 37 deg = 307 deg.)
The weight is 40 kg * 9.8 m/s^2 = 400 N, approx.. The components are therefore
wt_x = 400 N cos(233 deg) = -240 N and
wt_y = 400 N sin(233 deg) = -320 N
(if your ramp is inclined down and to the right your angle is 307 deg and the x component will be positive; the y component will be the same).
The only forces acting in the y direction are the y component of the weight and the normal force. Since we have y equilibrium (no acceleration in the y direction so zero net force in the y direction), these forces must be equal and opposite:
F_normal + wt_y = 0 so
F_normal = -wt_y = - ( -320 N) = 320 N.
The coefficient of friction is mu = .5. So the maximum possible frictional force has magnitude
| f_frict_max | = .5 * F_normal = .5 * 320 N = 160 N.
The x component of the weight, which acts parallel to the incline, has magnitude 240 N. Frictional force, which cannot exert more that 160 N, acts in the direction this force. So the net force is in the x direction, down the incline, and has magnitude
| F_net | = 240 N - 160 N = 80 N.
The net force is therefore 80 N down the incline.
The acceleration of the block is therefore
a = F_net / m = 80 N / (40 kg) = 2 m/s^2, down the incline.
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`q010. In the preceding, what coefficient of friction would be required for friction alone to hold the block stationary?
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To hold the block stationary would require a frictional force of 240 N, equal and opposite to the 'parallel component' of the weight.
The normal force is 320 N, and the frictional force cannot be greater than mu * F_normal. Therefore
mu * F_normal >= 240 N and
mu >= 240 N / (F_normal) = 240 N / (320 N) = .75.
The coefficient of friction must be at least .75.
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`q011. Continuing the same situation, suppose the block is sliding up the incline, subject only to gravitational, normal and frictional forces. What would be the acceleration of the block?
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If the block is sliding up the incline then the frictional force (160 N) is acting down the incline, in the same direction as the 'parallel component' of the weight (240 N).
So the magnitude of the net force will be
| F_net | = 160 N + 240 N = 400 N.
The acceleration of the system therefore has magnitude
| a | = | F_net | = 400 N / (40 kg) = 10 m/s^2.
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